I think we can show by induction that there exists such set B for every n >= 6.

It is enough to say $n\geq 5$. The structure of the set $A$ is a red herring; assume we have a finite set $A_0$ of positive integers such that $\sigma_0 = \sum_{x\in A_0} x^2 < \prod_{x\in A_0} x = \pi_0$ (here is when we need $n\geq 5$ in our problem, since for $n=4$ we have $1^2+2^2+3^2+4^2 = 30 > 4! = 24$). Take $A_1 = A_0 \cup \{\pi_0 - 1\}$, where obviously $\pi_0 - 1 \not \in A_0$. Then for $\sigma_1 = \sum_{x\in A_1} x^2 = \sigma_0 + (\pi_0 - 1)^2$ and $\pi_1 = \prod_{x\in A_1} x = \pi_0(\pi_0 - 1)$ we have $\pi_1 - \sigma_1 = \pi_0 - \sigma_0 - 1$. Then we repeat the procedure, for $k= \pi_0 - \sigma_0$ steps in all, until we get $ \pi_k - \sigma_k = 0$.

Actually the condition on $A$ is obviously unnecessary, any finite subset of positive integers works. And yes, I think the solution above works. It probably is the official one of the competition.

Indeed, if the finite set $A_0$ of positive integers is such that $\sigma_0 = \sum_{x\in A_0} x^2 \geq \prod_{x\in A_0} x = \pi_0$, we can work with $A_0\subset A'_0 = \{1,2,\ldots,n\}$, where $n = \max\{5, \max_{x\in A_0} x\}$, so $\sigma'_0 = \sum_{x\in A'_0} x^2 < \prod_{x\in A'_0} x = \pi'_0$.

mavropnevma wrote: It is enough to say $n\geq 5$. The structure of the set $A$ is a red herring; assume we have a finite set $A_0$ of positive integers such that $\sigma_0 = \sum_{x\in A_0} x^2 < \prod_{x\in A_0} x = \pi_0$ (here is when we need $n\geq 5$ in our problem, since for $n=4$ we have $1^2+2^2+3^2+4^2 = 30 > 4! = 24$). Take $A_1 = A_0 \cup \{\pi_0 - 1\}$, where obviously $\pi_0 - 1 \not \in A_0$. Then for $\sigma_1 = \sum_{x\in A_1} x^2 = \sigma_0 + (\pi_0 - 1)^2$ and $\pi_1 = \prod_{x\in A_1} x = \pi_0(\pi_0 - 1)$ we have $\pi_1 - \sigma_1 = \pi_0 - \sigma_0 - 1$. Then we repeat the procedure, for $k= \pi_0 - \sigma_0$ steps in all, until we get $ \pi_k - \sigma_k = 0$. Nice invariant.

Recently I encountered the following related problem (the source is unknown to me): Prove that there exists an integer $m \ge 2002$ and $m$ distinct positive integers $a_1, a_2, . . . , a_m$ such that $\prod a_i^2 - 4\sum\limits a_i^2$ is a perfect square. Using mavropnevma's procedure, we can obtain arbitrarily many distinct positive integers such that $\prod a_i = \sum\limits a_i^2 + 1$, which solves the problem because $\prod a_i^2 - 4\sum\limits a_i^2 = (\sum\limits a_i^2 + 1)^2 - 4\sum\limits a_i^2 = (\sum\limits a_i^2 - 1)^2$.

Seriously??

I mean the integers don't have to be distinct so technically we could append a bunch of $1$'s to $B$ so the product stays still but the sum increases. That is enough to finish the cases $n \ge 5$, while the rest can be solved by appending $n+1, ..., 5$ to $B$ such that the product surpasses the sum, and do the same. Someone correct me if I'm wrong

baenanabread wrote: I mean the integers don't have to be distinct so technically we could append a bunch of $1$'s to $B$ so the product stays still but the sum increases. That is enough to finish the cases $n \ge 5$, while the rest can be solved by appending $n+1, ..., 5$ to $B$ such that the product surpasses the sum, and do the same. Someone correct me if I'm wrong You are wrong. Recall that a set doesn't admit the same element more than once unless clearly stated (in which case it would be a multiset).